Question: $ A = \left[\begin{array}{rrr}0 & 2 & 2 \\ 0 & -1 & 0\end{array}\right]$ $ E = \left[\begin{array}{rr}2 & 3 \\ 0 & 4 \\ -1 & 2\end{array}\right]$ What is $ A E$ ?
Answer: Because $ A$ has dimensions $(2\times3)$ and $ E$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ A E = \left[\begin{array}{rrr}{0} & {2} & {2} \\ {0} & {-1} & {0}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{3} \\ {0} & \color{#DF0030}{4} \\ {-1} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{0}\cdot{2}+{2}\cdot{0}+{2}\cdot{-1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{2}+{2}\cdot{0}+{2}\cdot{-1} & ? \\ {0}\cdot{2}+{-1}\cdot{0}+{0}\cdot{-1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{2}+{2}\cdot{0}+{2}\cdot{-1} & {0}\cdot\color{#DF0030}{3}+{2}\cdot\color{#DF0030}{4}+{2}\cdot\color{#DF0030}{2} \\ {0}\cdot{2}+{-1}\cdot{0}+{0}\cdot{-1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{0}\cdot{2}+{2}\cdot{0}+{2}\cdot{-1} & {0}\cdot\color{#DF0030}{3}+{2}\cdot\color{#DF0030}{4}+{2}\cdot\color{#DF0030}{2} \\ {0}\cdot{2}+{-1}\cdot{0}+{0}\cdot{-1} & {0}\cdot\color{#DF0030}{3}+{-1}\cdot\color{#DF0030}{4}+{0}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-2 & 12 \\ 0 & -4\end{array}\right] $